{"id":48997,"date":"2019-11-14T02:33:10","date_gmt":"2019-11-14T01:33:10","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/o-que-e-o-calculo-da-cabeca-da-bomba-calculo-do-desempenho-da-bomba-definicao\/"},"modified":"2020-02-01T15:56:14","modified_gmt":"2020-02-01T14:56:14","slug":"o-que-e-o-calculo-da-cabeca-da-bomba-calculo-do-desempenho-da-bomba-definicao","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-o-calculo-da-cabeca-da-bomba-calculo-do-desempenho-da-bomba-definicao\/","title":{"rendered":"O que \u00e9 o c\u00e1lculo da cabe\u00e7a da bomba &#8211; C\u00e1lculo do desempenho da bomba &#8211; Defini\u00e7\u00e3o"},"content":{"rendered":"<div class=\"su-quote su-quote-style-default\">\n<div class=\"su-quote-inner su-clearfix\">Neste exemplo, veremos como prever a cabe\u00e7a da bomba e outras caracter\u00edsticas de desempenho da bomba. C\u00e1lculo da cabe\u00e7a da bomba &#8211; C\u00e1lculo do desempenho da bomba<\/div>\n<\/div>\n<div><\/div>\n<div><\/div>\n<div>\n<h2><span>C\u00e1lculo da cabe\u00e7a da bomba &#8211; C\u00e1lculo do desempenho da bomba<\/span><\/h2>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Pump-head-calculation.png\"><img loading=\"lazy\" class=\"alignright size-full wp-image-14958 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Pump-head-calculation.png\" alt=\"C\u00e1lculo da cabe\u00e7a da bomba\" width=\"264\" height=\"580\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Pump-head-calculation.png\" \/><\/a><span>Neste exemplo, veremos como prever<\/span><\/p>\n<ul>\n<li><strong><span>a descarga do projeto<\/span><\/strong><\/li>\n<li><a title=\"Pot\u00eancia de freio\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/centrifugal-pumps\/water-brake-horsepower\/\"><strong><span>cavalos-vapor de \u00e1gua<\/span><\/strong><\/a><\/li>\n<li><strong><span>a\u00a0<a title=\"Pump Head \u2013 Performance Curve of Centrifugal Pump\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/centrifugal-pumps\/pump-head-performance-curve\/\">cabe\u00e7a<\/a>\u00a0da\u00a0<\/span><a title=\"Cabe\u00e7a da bomba - curva de desempenho da bomba centr\u00edfuga\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/centrifugal-pumps\/pump-head-performance-curve\/\"><span>bomba<\/span><\/a><\/strong><\/li>\n<\/ul>\n<p><span>de uma\u00a0<\/span><a title=\"Bombas centr\u00edfugas\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-bomba-centrifuga-definicao\/\"><span>bomba centr\u00edfuga<\/span><\/a><span>\u00a0.\u00a0Esses dados de desempenho ser\u00e3o derivados da\u00a0<\/span><strong><span>equa\u00e7\u00e3o de turbom\u00e1quina de Euler:<\/span><\/strong><\/p>\n<p><strong><span>Torque do eixo: T\u00a0<\/span><sub><span>eixo<\/span><\/sub><span>\u00a0\u00a0 \u00a0 = \u03c1Q (r\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t2<\/span><\/sub><span>\u00a0&#8211; r\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t1<\/span><\/sub><span>\u00a0)<\/span><\/strong><\/p>\n<p><strong><span>Cavalos-for\u00e7a da \u00e1gua: P\u00a0<\/span><sub><span>w<\/span><\/sub><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = \u03c9.\u00a0<\/span><sub><span>Eixo \u00a0 \u00a0 \u00a0\u00a0<\/span><\/sub><span>\u00a0T\u00a0= \u03c1Q (u\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t2<\/span><\/sub><span>\u00a0&#8211; u\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t1<\/span><\/sub><span>\u00a0)<\/span><\/strong><\/p>\n<p><strong><span>Cabe\u00e7a da bomba: H = P\u00a0<\/span><sub><span>w<\/span><\/sub><span>\u00a0\/ \u03c1gQ = (u\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t2<\/span><\/sub><span>\u00a0&#8211; u\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t1<\/span><\/sub><span>\u00a0) \/ g<\/span><\/strong><\/p>\n<p><span>S\u00e3o fornecidos os seguintes dados para uma bomba de \u00e1gua centr\u00edfuga:<\/span><\/p>\n<ul>\n<li><strong><span>di\u00e2metros do impulsor<\/span><\/strong><span>\u00a0na entrada e na sa\u00edda<\/span>\n<ul>\n<li><strong><span>r\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0= 10 cm<\/span><\/strong><\/li>\n<li><strong><span>r\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= 20 cm<\/span><\/strong><\/li>\n<\/ul>\n<\/li>\n<li><strong><span>Velocidade = 1500 rpm<\/span><\/strong><span>\u00a0(rota\u00e7\u00f5es por minuto)<\/span><\/li>\n<li><span>o \u00e2ngulo da l\u00e2mina na entrada\u00a0<\/span><strong><span>\u03b2\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0= 30 \u00b0<\/span><\/strong><\/li>\n<li><span>o \u00e2ngulo da l\u00e2mina na sa\u00edda\u00a0<\/span><strong><span>\u03b2\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= 20 \u00b0<\/span><\/strong><\/li>\n<li><span>suponha que as larguras da l\u00e2mina na entrada e na sa\u00edda sejam:\u00a0<\/span><strong><span>b\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0= b\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= 4 cm<\/span><\/strong><span>\u00a0.<\/span><\/li>\n<\/ul>\n<p><span>Solu\u00e7\u00e3o:<\/span><\/p>\n<p><span>Primeiro, temos que calcular a\u00a0<\/span><strong><span>velocidade radial do fluxo<\/span><\/strong><span>\u00a0na sa\u00edda.\u00a0No diagrama de velocidade, a velocidade radial \u00e9 igual a (assumimos que o fluxo entra exatamente normal ao impulsor, portanto o componente tangencial da velocidade \u00e9 zero):<\/span><\/p>\n<p><strong><span>V\u00a0<\/span><sub><span>r1<\/span><\/sub><\/strong><span>\u00a0= u\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0tan 30 \u00b0 = \u03c9 r\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0tan 30 \u00b0 = 2\u03c0 x (1500\/60) x 0,1 x tan 30 \u00b0 =\u00a0<\/span><strong><span>9,1 m \/ s<\/span><\/strong><\/p>\n<p><span>O componente radial da velocidade de fluxo determina quanto a\u00a0<\/span><strong><span>taxa de fluxo de volume est\u00e1 entrando no impulsor<\/span><\/strong><span>\u00a0.\u00a0Portanto, quando conhecemos\u00a0<\/span><strong><span>V\u00a0<\/span><sub><span>r1<\/span><\/sub><\/strong><span>\u00a0na entrada, podemos determinar\u00a0<\/span><strong><span>a descarga<\/span><\/strong><span>\u00a0desta bomba de acordo com a seguinte equa\u00e7\u00e3o.\u00a0Aqui b\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0significa a largura da p\u00e1 do impulsor na entrada.<\/span><\/p>\n<p><strong><span>Q<\/span><\/strong><span>\u00a0=\u00a0<\/span><strong><span>2\u03c0.r\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0.b\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0.V\u00a0<\/span><sub><span>r1<\/span><\/sub>\u00a0<\/strong><span>= 2\u03c0 x 0,1 x 0,04 x 9,1 =\u00a0<\/span><strong><span>0,229 m\u00a0<\/span><sup><span>3<\/span><\/sup><span>\u00a0\/ s<\/span><\/strong><\/p>\n<p><span>Para calcular a\u00a0<\/span><strong><span>pot\u00eancia de \u00e1gua (P\u00a0<\/span><sub><span>w<\/span><\/sub><span>\u00a0)<\/span><\/strong><span>\u00a0necess\u00e1ria, temos que determinar a\u00a0<\/span><strong><span>velocidade do fluxo tangencial de sa\u00edda V\u00a0<\/span><sub><span>t2<\/span><\/sub><\/strong><span>\u00a0, porque foi assumido que a velocidade tangencial de entrada V\u00a0<\/span><sub><span>t1<\/span><\/sub><span>\u00a0\u00e9 igual a zero.<\/span><\/p>\n<p><span>A velocidade do fluxo radial de sa\u00edda segue da\u00a0<\/span><strong><span>conserva\u00e7\u00e3o de Q<\/span><\/strong><span>\u00a0:<\/span><\/p>\n<p><strong><span>Q = 2\u03c0.r\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0.b\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0.V\u00a0<\/span><sub><span>r2<\/span><\/sub><\/strong><span>\u00a0\u00a0\u21d2\u00a0<\/span><strong><span>V\u00a0<\/span><\/strong><strong><sub><span>r2<\/span><\/sub><\/strong><span>\u00a0= Q \/ 2\u03c0.r\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0.b\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= 0,229 \/ (2\u03c0 x 0,2 x 0,04) =\u00a0<\/span><strong><span>\u00a04,56 m \/ s<\/span><\/strong><\/p>\n<p><span>A partir do\u00a0\u00e2ngulo da l\u00e2mina de sa\u00edda\u00a0da figura (\u00a0<\/span><strong><span>tri\u00e2ngulo da velocidade<\/span><\/strong><span>\u00a0), \u03b2\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0, pode ser facilmente representado da seguinte maneira.<\/span><\/p>\n<p><strong><span>ber\u00e7o \u03b2\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= (u\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0&#8211; V\u00a0<\/span><sub><span>t2<\/span><\/sub><span>\u00a0) \/ V\u00a0<\/span><sub><span>r2<\/span><\/sub><\/strong><\/p>\n<p><span>e, portanto, a velocidade do fluxo tangencial de sa\u00edda V\u00a0<\/span><sub><span>t2<\/span><\/sub><span>\u00a0\u00e9:<\/span><\/p>\n<p><strong><span>V\u00a0<\/span><sub><span>t2<\/span><\/sub><\/strong><span>\u00a0=\u00a0<\/span><strong><span>u\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0&#8211; V\u00a0<\/span><sub><span>r2<\/span><\/sub><span>\u00a0.\u00a0ber\u00e7o 20 \u00b0<\/span><\/strong><span>\u00a0= \u03c9 r\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0&#8211; V\u00a0<\/span><sub><span>R2<\/span><\/sub><span>\u00a0.\u00a0ber\u00e7o 20 \u00b0 = 2\u03c0 x 1500\/60 x 0,2 &#8211; 4,56 x 2,75 = 31,4 &#8211; 12,5 =\u00a0<\/span><strong><span>18,9 m \/ s.<\/span><\/strong><\/p>\n<p><span>A pot\u00eancia de \u00e1gua necess\u00e1ria \u00e9 ent\u00e3o:<\/span><\/p>\n<p><strong><span>P\u00a0<\/span><sub><span>w<\/span><\/sub><span>\u00a0\u00a0= \u03c1 Q u\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t2<\/span><\/sub>\u00a0<\/strong><span>= 1000 [kg \/ m\u00a0<\/span><sup><span>3<\/span><\/sup><span>\u00a0] x 0,229 [m\u00a0<\/span><sup><span>3<\/span><\/sup><span>\u00a0\/ s] x 31,4 [m \/ s] x 18,9 [m \/ s] = 135900 W =\u00a0<\/span><strong><span>135,6 kW<\/span><\/strong><\/p>\n<p><span>e a cabe\u00e7a da bomba \u00e9:<\/span><\/p>\n<p><strong><span>H \u2248 P\u00a0<\/span><sub><span>w<\/span><\/sub><span>\u00a0\/ (\u03c1 g Q)<\/span><\/strong><span>\u00a0= 135900 \/ (1000 x 9,81 x 0,229) =\u00a0<\/span><strong><span>60,5 m<\/span><\/strong><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"lgc-column lgc-grid-parent lgc-grid-100 lgc-tablet-grid-100 lgc-mobile-grid-100 lgc-equal-heights lgc-first lgc-last\">\n<div class=\"inside-grid-column\">\n<div class=\"su-accordion\">\n<p>&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.<\/p>\n<p>Este artigo \u00e9 baseado na tradu\u00e7\u00e3o autom\u00e1tica do artigo original em ingl\u00eas. Para mais informa\u00e7\u00f5es, consulte o artigo em ingl\u00eas. Voc\u00ea pode nos ajudar. Se voc\u00ea deseja corrigir a tradu\u00e7\u00e3o, envie-a para: translations@nuclear-power.com ou preencha o formul\u00e1rio de tradu\u00e7\u00e3o on-line. Agradecemos sua ajuda, atualizaremos a tradu\u00e7\u00e3o o mais r\u00e1pido poss\u00edvel. Obrigado.<\/p>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Neste exemplo, veremos como prever a cabe\u00e7a da bomba e outras caracter\u00edsticas de desempenho da bomba. C\u00e1lculo da cabe\u00e7a da bomba &#8211; C\u00e1lculo do desempenho da bomba C\u00e1lculo da cabe\u00e7a da bomba &#8211; C\u00e1lculo do desempenho da bomba Neste exemplo, veremos como prever a descarga do projeto cavalos-vapor de \u00e1gua a\u00a0cabe\u00e7a\u00a0da\u00a0bomba de uma\u00a0bomba centr\u00edfuga\u00a0.\u00a0Esses dados &#8230; <a title=\"O que \u00e9 o c\u00e1lculo da cabe\u00e7a da bomba &#8211; C\u00e1lculo do desempenho da bomba &#8211; Defini\u00e7\u00e3o\" class=\"read-more\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-o-calculo-da-cabeca-da-bomba-calculo-do-desempenho-da-bomba-definicao\/\" aria-label=\"More on O que \u00e9 o c\u00e1lculo da cabe\u00e7a da bomba &#8211; C\u00e1lculo do desempenho da bomba &#8211; Defini\u00e7\u00e3o\">Ler mais<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[14],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v15.4 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>O que \u00e9 o c\u00e1lculo da cabe\u00e7a da bomba - C\u00e1lculo do desempenho da bomba - Defini\u00e7\u00e3o<\/title>\n<meta name=\"description\" content=\"Neste exemplo, veremos como prever a cabe\u00e7a da bomba e outras caracter\u00edsticas de desempenho da bomba. 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